If $x,y \in (0,\frac{\pi}{2})$ then expression $\sin x +\cos y +\tan^2y+\cot^2x+5>\ldots?$ Problem : 
If $x,y \in (0,\frac{\pi}{2})$ then expression $\sin x +\cos y +\tan^2y+\cot^2x+5$ is always greater than : 
(a)  $\ 7 $
(b)  $\ 8 $
(c)  $\ 9 $
(d)  $\ $none of these 
Solution :
We can write the given expression $\sin x +\cos y +\tan^2y+\cot^2x+5$ as $\sin x +\cot^2x +\cos x +\tan^2y +5$ 
$\Rightarrow \sin x +\cot^2x +\cos y +\tan^2y +5  = \sin x + \csc^2x -1 +\cos x +\sec^2y-1+5$
$\Rightarrow \sin x + \csc^2x  +\cos x +\sec^2y+3$ 
Please guide me how to proceed further. Thanks.
 A: I think you are doing a mistake here. 
$ \sin(x)+\cos(y)+\tan^2(y)+\cot^2(x)+5 = (\sin(x)+ \csc^2(x)) + (\cos(y)+ 
\sec^2(y))+ 3$ . Not, what you have written. 
Considering that to be a typo, Notice that the things in the bracket are independent, so we just find minimum value of each  separately.
Consider $f(x)=\sin(x)+\csc^2(x) \ge \sin(x)+\csc(x) \ge 2$.
(By AM-GM inequality) . Similarly consider $g(x)=\cos(y)+\sec^2(y)$
We get $7$ as the answer .
A: I don't think the simplifying is necessary.  We can just analyze the functions within the open interval $(0,\pi/2)$.  First, we can examine the minimum value of $\sin x+\cot^2x$.  When $x=\pi/2$, sine is $1$ and cotangent is $0$.  Cotangent increases without bound as we move towards $0$ from $\pi/2$.  So, we can assume $\sin x+\cot^2x$ is always greater than $1$ on the interval $(0,\pi/2)$.  The same reasoning applies to $\cos y+\tan^2 y$, however the minimum will be at $y=0$ now, and the minimum is again $1$.  So, just add these together with $5$, and we see that the entire function is always greater than $7$.
